Question

一题数学题，求解。

a+b=1,
a^2+b^2=3
,a^3+b^3=4,
a^4+b^4=7,
a^5+b^5=11.
求a^10+b^10..
过程......

Answer 1

答案是：a^10+b^10=123
具体解答步骤：a+b=1能推出（a+b）^2=1推出ab=-1
a^6+b^6=(a^5+b^5)(a+b)-ab(a^4+b^4)=18
依次可得：a^7+b^7=(a^6+b^6)(a+b)-ab(a^5+b^5)=29
a^8+b^8=(a^7+b^7)(a+b)-ab(a^6+b^6)=47
a^9+b^9=(a^8+b^8)(a+b)-ab(a^7+b^7)=76
a^10+b^10=(a^9+b^9)(a+b)-ab(a^8+b^8)=123
由此规律可知：a^n+b^n=(a^n-1+b^n-1）+(a^n-2+b^n-2)

Answer 2

a+b=1,
a^2+b^2=3
ab=(1/2)[(a+b)^2-(a^2+b^2)]=(1/2)(1-3)=-1
a^5+b^5=11.
a^10+b^10=(a^5+b^5)^2-2a^5b^5=11×11-2×（-1）^5=121+2=123

Answer 3

由(a+b)^2=1得ab=-1
(a^5+b^5)^2-2a^5b^5=11^2-2*(-1)即121+2=123

Answer 4

a+b)^2=a^2+b^2+2ab
ab=-1
a^10+b^10=(a^5+b^5)^2-2a^5*b^5
=11^2-2*(-1)^5
=123

Answer 5

(a+b)^2=a^2+b^2+2ab
ab=-1
a^10+b^10=(a^5+b^5)^2-2a^5*b^5
=11^2-2*(-1)^5
=123